视图控制器之间不传递的数据无法解决

Question

我试图在两个视图控制器之间传递一个字符串，第一个名为 DetailController，第二个名为 VideoController。它们使用 push segue 与按钮连接。正在调用我的“prepareForSegue”方法，但是该字符串在 VideoController 中返回为 null。我被这个问题困扰了很久，仍然找不到解决方案。任何帮助将不胜感激。

DetailController.m prepareForSegue 方法

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{   NSString *myString=self.city.camera;
    NSLog(@"%@", myString);
    NSString *see=@"what is going on?";
    NSLog(@"%@", see);

    VideoController *video= (VideoController *) segue.destinationViewController;
    video.cam= self.city.camera;

}

视频控制器.m

#import "VideoController.h"
#import "City.h"
#import "DetailController.h"
#import "ViewController.h"

@interface VideoController ()

@end

@implementation VideoController
@synthesize webView,city,cam;



- (void)viewDidLoad
{
    NSString *see = self.cam;
    NSLog(@"%@", see);

    [super viewDidLoad];
    // Do any additional setup after loading the view.

    [self performSelector:@selector(yourMethod) withObject:nil afterDelay:10.0];





    NSString * html = [NSString stringWithFormat:@"<img name=\"Cam\" src=\"%@\" width=\"110%%\"  height=\"100%%\" alt=\"Live Feed\" style=\"background-color: #000000\" />", cam];

    [webView loadHTMLString:html baseURL:nil];

}
-(void)yourMethod
{
    [self.navigationController popViewControllerAnimated:YES];
}

- (void)didReceiveMemoryWarning
{
    [super didReceiveMemoryWarning];
    // Dispose of any resources that can be recreated.
}

@end

这是我在按下按钮并发生转场时收到的消息。如果我正确阅读它，当它应该来自不同的webView时，（null）将由与其他两个字符串相同的webView返回？我完全被这个问题迷失了

2014-04-21 02:20:23.093 webView[47963:1a303] http://admin:*******@192.168.1.80/video.cgi/
2014-04-21 02:20:23.094 webView[47963:1a303] what is going on?
2014-04-21 02:20:23.102 webView[47963:1a303] (null)

这是我的 VideoController.h 文件，您可以看到 sam 设置为 strong，我是否遗漏了什么？

#import <UIKit/UIKit.h>
#import "DetailController.h"
@class City;
@interface VideoController : UIViewController{
IBOutlet UIWebView *webView;
IBOutlet UIImageView *image;



}
@property (strong,nonatomic) City *city;
@property (strong, nonatomic)NSString * cam;
@property(nonatomic,retain) UIWebView *webView;


@end

Answer 1

看我的评论：

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{   
   NSString *myString=self.city.camera;
   NSLog(@"%@", myString);
   NSString *see=@"what is going on?";
   NSLog(@"%@", see);

   VideoController *video= (VideoController *) segue.destinationViewController;
   // be sure that self.city.camera is has value 
   // I think you assign nil to video.cam
   video.cam= self.city.camera;
}

Answer 2

确保您的“cam”属性没有分配弱属性，否则它可能会在您需要之前被释放。更好的是，发布您的 VideoController.h 代码。